please help me with calculus?

Question

its multiple choice, but i need work or an explanation...

let f be a function defined as follows, where a does not = 0.

f(x)= (x^2 - a^2) / (x-a) for x does not equal a

f(x)=0 for x equals a

which of the following are true about f(x)?

1. limit as x goes to a exists
2. f(a) exists
3. f(x) is continuous at x=a

a. none
b. 1 only
c. 2 only
d. 1 and 2 only
e. 1, 2, and 3

Answer 1

Many wrong answers here.
First, f(a) is *defined* to be 0, so f(a) exists.
Next, lim f(x)=lim (x-a)(x+a)/(x-a)=lim x+a=2a, so the limit exists.
Now, since you are given that a is not 0, the value of f(a) is not the same as the limit of f(x) as x goes to a, so f is *not* continuous at x=a.
The answer is D.

Answer 2

for a limit to exist, it must be the same from both directions (i.e lim x->a+=lim x->a-)
this function satisfies this requirement so (1) is correct.

for f(a) to exist, it must have some value, which in this case is undefined, regardless of what we put in for a, so clearly it does NOT satisfy (3) since to be continuous at a, the function must equal the limit of x->a at a

and finally f(x)=0 when x=a is given so f(a) does exist. therefore answer d is right

Answer 3

the answer is B only the first statement is true
F(a) does not exist because the denominator of a fraction in a function cannot equal zero. it is undefined, not zero, take a graphing calculator and try to graph x/0 = y you will not get a graph

F(x) is not continuous because f(a) does not exist, there is a hole in the function

the second answerer is right , but he is also wrong. what he said is correct, but f(x) is still not continous, if you graph the original function you still have a hole at f(a)

Answer 4

2 only

Answer 5

Does it help any if you notice that

(x^2 - a^2) = (x - a)(x + a) ?

and dividing by (x - a) will leave you with f(x) = x + a ?